\(R= \sqrt{2+\sqrt{3}}. \sqrt{2+\sqrt{2 +\sqrt{3}}}. \sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
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13 tháng 6 2023
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\sqrt{4-2-\sqrt{2+\sqrt{3}}}\)
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{4-2-\sqrt{3}}\)
=1
13 tháng 6 2023
\(=\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}}\sqrt{2-\sqrt{2+\sqrt{3}}}\)
\(=\sqrt{2+\sqrt{3}}\sqrt{2-\sqrt{3}}=1\)
24 tháng 5 2022
e có phát hiện mới:v cj chung lớp vs cj kia đúng hemm:v
VD
25 tháng 5 2022
\(R=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\left(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\right)\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{\left(2+\sqrt{2+\sqrt{2+\sqrt{3}}}\right)\left(2-\sqrt{2+\sqrt{2+\sqrt{3}}}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2^2-\left(\sqrt{2+\sqrt{2+\sqrt{3}}}\right)^2}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{4-\left(2+\sqrt{2+\sqrt{3}}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\)
\(=\sqrt{2+\sqrt{3}}.\left(\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\right)\\ =\sqrt{2+\sqrt{3}}.\sqrt{\left(2+\sqrt{2+\sqrt{3}}\right)\left(2-\sqrt{2+\sqrt{3}}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2^2-\left(\sqrt{2+\sqrt{3}}\right)^2}\\ =\sqrt{2+\sqrt{3}}.\sqrt{4-\left(2+\sqrt{3}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}\\ =\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\\ =\sqrt{4-\sqrt{3^2}}\\ =\sqrt{4-3}\\ =\sqrt{1}\\ =1\)
AH
Akai Haruma
Giáo viên
28 tháng 7 2018
Lời giải:
Đặt \(\sqrt{2+\sqrt{3}+\sqrt{2-\sqrt{3}}}=a; \sqrt{2+\sqrt{3}-\sqrt{2-\sqrt{3}}}=b\)
Có:
\(a^2+b^2=(2+\sqrt{3}+\sqrt{2-\sqrt{3}})+(2+\sqrt{3}-\sqrt{2-\sqrt{3}})=2(2+\sqrt{3})\)
\(=4+2\sqrt{3}=3+1+2\sqrt{3.1}=(\sqrt{3}+1)^2\)
\(ab=\sqrt{(2+\sqrt{3}+\sqrt{2-\sqrt{3}})(2+\sqrt{3}-\sqrt{2-\sqrt{3}})}\)
\(=\sqrt{(2+\sqrt{3})^2-(2-\sqrt{3})}=\sqrt{5+5\sqrt{3}}\)
Như vậy:
\(\frac{\sqrt{2+\sqrt{3}+\sqrt{2-\sqrt{3}}}}{\sqrt{2+\sqrt{3}-\sqrt{2-\sqrt{3}}}}+\frac{\sqrt{2+\sqrt{3}-\sqrt{2-\sqrt{3}}}}{\sqrt{2+\sqrt{3}+\sqrt{2-\sqrt{3}}}}=\frac{a}{b}+\frac{b}{a}=\frac{a^2+b^2}{ab}\)
\(=\frac{(\sqrt{3}+1)^2}{\sqrt{5+5\sqrt{3}}}=\frac{(\sqrt{3}+1)^2}{\sqrt{5}.\sqrt{\sqrt{3}+1}}=\frac{(\sqrt{3}+1)^{1.5}}{\sqrt{5}}\)
N
19 tháng 9 2016
\(\sqrt{2+\sqrt{3}.}\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{\left(2+\sqrt{2+\sqrt{2+\sqrt{3}}}\right)\left(2-\sqrt{2+\sqrt{2+\sqrt{3}}}\right)}\)
=\(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{4-2-\sqrt{2+\sqrt{3}}}\)
=\(\sqrt{2+\sqrt{3}}.\sqrt{\left(2+\sqrt{2+\sqrt{3}}\right).\left(2-\sqrt{2+\sqrt{3}}\right)}\)
=\(\sqrt{2+\sqrt{3}}.\sqrt{4-2-\sqrt{3}}=\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}\)
=\(\sqrt{4-3}=1\)
học ghê v, đến tận đây r à :V
\(R=\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\left(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\right)\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{\left(2+\sqrt{2+\sqrt{2+\sqrt{3}}}\right)\left(2-\sqrt{2+\sqrt{2+\sqrt{3}}}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2^2-\left(\sqrt{2+\sqrt{2+\sqrt{3}}}\right)^2}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{4-\left(2+\sqrt{2+\sqrt{3}}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\)
\(=\sqrt{2+\sqrt{3}}.\left(\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2-\sqrt{2+\sqrt{3}}}\right)\\ =\sqrt{2+\sqrt{3}}.\sqrt{\left(2+\sqrt{2+\sqrt{3}}\right)\left(2-\sqrt{2+\sqrt{3}}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2^2-\left(\sqrt{2+\sqrt{3}}\right)^2}\\ =\sqrt{2+\sqrt{3}}.\sqrt{4-\left(2+\sqrt{3}\right)}\\ =\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}\\ =\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\\ =\sqrt{4-\sqrt{3^2}}\\ =\sqrt{4-3}\\ =\sqrt{1}\\ =1\)